Solve for $n$, $- \dfrac{3n + 4}{4n + 1} = \dfrac{1}{3} $
Solution: Multiply both sides of the equation by $4n + 1$ $ -(3n + 4) = \dfrac{4n + 1}{3} $ Multiply both sides of the equation by $3$ $ -3(3n + 4) = 4n + 1 $ $-9n - 12 = 4n + 1$ $-12 = 13n + 1$ $-13 = 13n$ $13n = -13$ $n = -\dfrac{13}{13}$ Simplify. $n = -1$